An oversized yo-yo is made from two identical solid disks each of mass M = 1.90

Question

An oversized yo-yo is made from two identical solid disks each of mass

M = 1.90 kg

and radius

R = 11.9 cm.

The two disks are joined by a solid cylinder of radius

r = 4.00 cm

and mass

m = 1.00 kg

as in the figure below. Take the center of the cylinder as the axis of the system, with positive torques directed to the left along this axis. All torques and angular variables are to be calculated around this axis. Light string is wrapped around the cylinder, and the system is then allowed to drop from rest.

(a) What is the moment of inertia of the system? Give a symbolic answer. (Use any variable or symbol stated above as necessary. Do not substitute numerical values; use variables only.)
I =   

(b) What torque does gravity exert on the system with respect to the given axis?
N · m

(c) Take downward as the negative coordinate direction. As depicted in the figure above, is the torque exerted by the tension positive or negative?

positivenegative     

Is the angular acceleration positive or negative?

positivenegative     

What about the translational acceleration?

positivenegative     

(d) Write an equation for the angular acceleration (alpha) in terms of the translational acceleration a and radius r. (Watch the sign!)

(e) Write Newton's second law for the system in terms of m, M, a, T, and g. (Submit a file with a maximum size of 1 MB.)

This answer has not been graded yet.

(f) Write Newton's second law for rotation in terms of I, , T, and r. (Submit a file with a maximum size of 1 MB.)

This answer has not been graded yet.

(g) Eliminate from the rotational second law with the expression found in part (d) and find a symbolic expression for the acceleration a in terms of m, M, g, r and R.

(h) What is the numeric value for the system's acceleration?
m/s2

(i) What is the tension in the string?
N

(j) How long does it take the system to drop 1.03 m from rest?
s

Explanation / Answer

Here,

a)

as the axis of ratation is the centre axis of cyclinders

Total moment of inertia of system = 2 * moment of inertia of disk + moment of inertia of cyclinder

Total moment of inertia of system = 2 * 0.5 * M * R^2 + 0.5 * m * r^2

Total moment of inertia of system = 1.90 * 0.119^2 + 0.5 * 1 * 0.04^2

Total moment of inertia of system = 0.0277 Kg.m^2

b)

Torque applied by gravity = mg * 0

as the centre of gravity will lie on the axis of rotation

Torque applied by gravity = 0 N.m

c)

as the tension in upwards

Torque applied by the tension is positive

-------------

as Torque applied by the tension is positive

angular acceleration is positive

--------------

as the translational acceleration will be downwards

the translational acceleration is negative

d)

Here ,

for the accelerations

= - a/r

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